Problem: Simplify the following expression and state the condition under which the simplification is valid. $x = \dfrac{-9y^2 - 27y + 36}{y^3 - 2y^2 - 24y}$
First factor out the greatest common factors in the numerator and in the denominator. $ x = \dfrac {-9(y^2 + 3y - 4)} {y(y^2 - 2y - 24)} $ $ x = -\dfrac{9}{y} \cdot \dfrac{y^2 + 3y - 4}{y^2 - 2y - 24} $ Next factor the numerator and denominator. $ x = - \dfrac{9}{y} \cdot \dfrac{(y + 4)(y - 1)}{(y + 4)(y - 6)}$ Assuming $y \neq -4$ , we can cancel the $y + 4$ $ x = - \dfrac{9}{y} \cdot \dfrac{y - 1}{y - 6}$ Therefore: $ x = \dfrac{ -9(y - 1)}{ y(y - 6)}$, $y \neq -4$